1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
pshichka [43]
2 years ago
12

I don’t understand please help

Chemistry
2 answers:
dexar [7]2 years ago
6 0
It’s C)H2SO4
Good luck
lina2011 [118]2 years ago
4 0
I agree the answer is C according to my research about it
You might be interested in
Calculate AH°rxn from AH°7 values (use table in textbook appendix) a) Cl2 (g) + 2 Na (s) -› 2 NaCl (s) EITHER -411.1kJ/mol b) 2
MrMuchimi

To find AH°rxn, we use the following equation:

What we're going to do is to sum the enthalpy of the products and then substract with the enthalpy of the reactives:

As you can see, we need to multiply by the coefficients of the reaction.

Now, just replace the values of the table:

So the answer is -822.2kJ/mol.

For b:

Now, just replace the values of the table:

The answer for b is -1036kJ/mol.

4 0
1 year ago
The type of compound that is most likely to contain a covalent bond is ________.
max2010maxim [7]
The type of substance that is most likely to contain a covalent bond is ONE THAT IS COMPOSED OF ONLY NON METALS.
Covalent bond is a type of chemical bond in which electron pairs are shared among the participating elements in order to achieve the octet form. Covalent bond is usually found among non metals.
6 0
3 years ago
A river with a flow of 50 m3/s discharges into a lake with a volume of 15,000,000 m3. The river has a background pollutant conce
spin [16.1K]

Explanation:

The given data is as follows.

       Volume of lake = 15 \times 10^{6} m^{3} = 15 \times 10^{6} m^{3} \times \frac{10^{3} liter}{1 m^{3}}

        Concentration of lake = 5.6 mg/l

Total amount of pollutant present in lake = 5.6 \times 15 \times 10^{9} mg

                                                                    = 84 \times 10^{9} mg

                                                                    = 84 \times 10^{3} kg

Flow rate of river is 50 m^{3} sec^{-1}

Volume of water in 1 day = 50 \times 10^{3} \times 86400 liter

                                          = 432 \times 10^{7} liter

Concentration of river is calculated as 5.6 mg/l. Total amount of pollutants present in the lake are 2.9792 \times 10^{10} mg or 2.9792 \times 10^{4} kg

Flow rate of sewage = 0.7 m^{3} sec^{-1}

Volume of sewage water in 1 day = 6048 \times 10^{4} liter

Concentration of sewage = 300 mg/L

Total amount of pollutants = 1.8144 \times 10^{10} mg or 1.8144 \times 10^{4}kg

Therefore, total concentration of lake after 1 day = \frac{131936 \times 10^{6}}{1.938 \times 10^{10}}mg/ l

                                        = 6.8078 mg/l

                 k_{D} = 0.2 per day

       L_{o} = 6.8078

Hence, L_{liquid} = L_{o}(1 - e^{-k_{D}t}

             L_{liquid} = 6.8078 (1 - e^{-0.2 \times 1})  

                             = 1.234 mg/l

Hence, the remaining concentration = (6.8078 - 1.234) mg/l

                                                             = 5.6 mg/l

Thus, we can conclude that concentration leaving the lake one day after the pollutant is added is 5.6 mg/l.

5 0
3 years ago
Select all that apply. Catalysts can save money by essentially lowering the:
k0ka [10]

Answer: Options (a) and (d) are the correct answer.

Explanation:

A catalyst is the substance which helps in increasing the rate of reaction.

Activation energy is the minimum amount of energy required by reactants to start the reaction. On addition of catalyst, the path of reaction changes because the energy barrier gap reduces and hence, the activation energy also decreases.

In the absence of catalyst, we need to increase the temperature so that reaction can occur quickly.

Whereas on addition of catalyst, there is no need to increase the temperature as the catalyst itself is sufficient to increase the rate of reaction. As a result, temperature should be lowered when there is addition of catalyst in the reaction.

Thus, we can conclude that catalysts can save money by essentially lowering the activation energy and temperature required.


4 0
3 years ago
Read 2 more answers
Picture of gas laws question below:
Vera_Pavlovna [14]

Volume of the tank is 5.5 litres.

Explanation:

mass of the CO2 is given 8.6 grams

Pressure of the gas is 89 Kilopascal which is 0.8762 atm

Temperature of the gas is 29 degrees ( 0 degrees +273.5= K) so (29+273)

R = gas constant 0.0821 liter atmosphere per kelvin)

FROM THE IDEAL GAS LAW

PV=nRT ( P Pressure, V Volume, n is number of moles of gas, R gas constant, Temperature in Kelvin)

no of moles = mass/atomic mass

                    =  8.6/44

                    = 0.195 moles

now putting the values in equation

V=nRT/P

  = 0.195*0.0821*302/ 0.8762

  = 5.5 litres.

As the carbon dioxide gas occupies the volume os the tank hence volume of tank is 5.5 litres.

4 0
3 years ago
Other questions:
  • What is the one way the otter is adapted to living in the moving water of a river
    15·1 answer
  • What CPT code is reported for a percutaneous needle biopsy of mediastinum?
    13·2 answers
  • Rutherford's gold-foil experiments proved the existence of which of the following?
    7·2 answers
  • How to find percentage abundanace
    11·1 answer
  • What is the ph of with a 4.6 x 10-6 M hydroxide ion concentration?
    15·1 answer
  • What property has a higher value than silicon
    9·1 answer
  • Hydrogen gas and nitrogen gas are used to make ammonia (NH3). Calculate the rate of reaction if the concentration of hydrogen ch
    15·1 answer
  • Hydrogen and oxygen combine in the ratio of 1:8 by mass to form water. How many grams of hydrogen are needed to react completely
    12·1 answer
  • Explain how the surrounding water (the Atlantic Ocean and the Long Island Sound) affects the climate of Long Island. Make sure t
    12·1 answer
  • 3. What are the Earth's Poles? *
    9·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!